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This is a follow-up of this question.

Let $V,W$ be $d$-dimensional real vector spaces, and let $A,B \in \text{Hom}(V,W)$ be non-invertible maps, $1 \le k \le d-1$.

Consider the induced maps $\bigwedge^{k}A,\bigwedge^{k}B :\Lambda_k(V) \to \Lambda_k(W)$. I want to characterize all the pairs $(A,B)$ which satisfy $\bigwedge^k A=\bigwedge^k B \neq 0.$

Partial results:

$(1)$ I prove below that a necessary condition is $\ker A=\ker B, \text{Image} \, A=\text{Image} \, B$. In particular $\text{rank} \,A=\text{rank} \,B$.

$(2)$ Moreover, contrary to the case where $A,B$ are invertible when the only possible pairs are $(A,A),(A,-A)$, in the singular case there is much more freedom:

Take a $3$-dimensional space with a basis $v_1,v_2,v_3$, and define $A,B$ via: $Av_3=Bv_3=0,Av_1=2v_1,Av_2=\frac{1}{2}v_2,Bv_1=\frac{1}{2}v_1,Bv_2=2v_2$.

This is can be generalized into the following sufficient condition:

$A,B$ have identical eigenspaces, null spaces, and equal products of all the nonzero eigenvalues.

Essentially, since $\ker A=\ker B=W$, $\text{Image} \, A=\text{Image} \, B=\tilde W$, the question really seems about the invertible quotient operators:

$\tilde A,\tilde B:V/W \to \tilde W$.

(I think my example above is related to the special case where $k=\text{rank} \,B=\dim (\tilde W)$. The right condition thenseems to be the "determinant" of the quotient operator is $1$).


Proof that $\ker A=\ker B$:

Let $v \in \ker A$ and assume by contradiction that $Bv \neq 0$. Fix some inner product on $V$, in such a way that $v \in (\ker B)^{\perp}$. Since $\dim(\ker B)^{\perp} = \text{rank} {B} \ge k$, there exist $v_1,\dots,v_{k-1} \in (\ker B)^{\perp}$ such that $v,v_1,\dots,v_{k-1}$ are linearly independent.

Then $ 0=\bigwedge^k A(v \wedge v_1 \wedge \dots \wedge v_{k-1})= Bv \wedge Bv_1 \wedge \dots \wedge Bv_{k-1}$, hence $Bv,Bv_1,\dots,Bv_{k-1}$ are linearly dependent, in contradiction to the fact that $B|_{(\ker B)^{\perp}}$ is injective.

This shows $\ker A \subseteq \ker B$. The other direction follows from symmetry.

The proof that $\text{Image} \, A=\text{Image} \, B$ is easy.

*(I am sure there is a cleaner argument with no insertion of an inner product, e.g via quotients).

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  • $\begingroup$ Are you sure? I meant to assume equality for a single $k$, not all of them... $\endgroup$ – Asaf Shachar Jul 8 '17 at 15:44
  • $\begingroup$ The proof works for a single fixed $k$, at least to me. $\endgroup$ – Gunnar Þór Magnússon Jul 8 '17 at 16:00
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It seems to me that all the ingredients of a full answer are included in your question. I'll just put them in the right order.

Yes, $A$ and $B$ have the same kernel $U$ and the same image $\tilde{W}$, so we may think of the two invertible operators $$A',B':V/U\to\tilde{W}.$$

Now, you just need to distinguish between two cases. If $\dim\tilde{W}=k$, then a necessary and sufficient condition is that both operators have the same determinant (this is not even a condition, just another way to formulate the equality you are talking about).

If $\dim\tilde{W}>k,$ the answer to your previous question applies again. That is, $A'$ and $B'$ have to differ by a constant $\alpha$ satisfying $\alpha^k=1$.

It is interesting that the first case allows a big space of solutions (something like $\text{SL}(V/U))$, whereas the solutions in the second case are all trivial.

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  • $\begingroup$ Thanks. Somehow I feel that the notion of determinant of the quotient maps $A',B'$ shouldn't be well-defined, since in general the determinant of a linear map between two different vector spaces, is not well-defined (unless you add some additional structure such as metrics or preferred volume forms on these spaces). This was the reason for my hesitation in putting "determinant". For a map $V \to V$ (that is from a vector space to itself) the determinant is always defined without any need for more structure (just take some top-form $\omega$, $\endgroup$ – Asaf Shachar Jul 8 '17 at 20:45
  • $\begingroup$ then $\det T$ is defined by $T^*\omega =\det T \cdot \omega$. $\endgroup$ – Asaf Shachar Jul 8 '17 at 20:47
  • $\begingroup$ Nevertheless, @AsafShachar, saying that two different operators $V\to W$ have the same determinant is well defined, even with no extra structure. If you like, you can replace this by "$\det B^{-1}A=1$". $\endgroup$ – Amitai Yuval Jul 8 '17 at 21:23
  • $\begingroup$ @AsafShachar In other words, this just depends on your definition for the determinant of a linear map $V\to W$. My definition is the induced linear map of the top exterior powers. If $V=W$, this map can be thought of as a number. $\endgroup$ – Amitai Yuval Jul 8 '17 at 22:32

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